Don't use aggregate's simplify=TRUE when FUN() produces return
values of various dimensions. In your case, the shape of table(subset)'s
return value depends on the number of levels in the factor 'subset'.
If you make B a factor before splitting it by C, each split will have the
same number of levels (2). If you split it and then let table convert
each split to a factor, one split will have 1 level and the other 2.
To see
the details of the output , use str() instead of print().
Bill Dunlap
TIBCO Software
wdunlap tibco.com <http://tibco.com>
On Tue, Feb 6, 2018 at 12:20 AM, Alain Guillet
<alain.guil...@uclouvain.be <mailto:alain.guil...@uclouvain.be>> wrote:
Dear R users,
When I use aggregate with table as FUN, I get what I would call a
strange behaviour if it involves numerical vectors and one "level"
of it is not present for every "levels" of the "by" variable:
---------------------------
> df <-
data.frame(A=c(1,1,1,1,0,0,0,0),B=c(1,0,1,0,0,0,1,0),C=c(1,0,1,0,0,1,1,1))
> aggregate(df[1:2],list(df$C),table,simplify = TRUE,drop=TRUE)
Group.1 A.0 A.1 B
1 0 1 2 3
2 1 3 2 2, 3
> table(df$C,df$B)
0 1
0 3 0
1 2 3
---------------
As you can see, a comma appears in the column with the variable B
in the aggregate whereas when I call table I obtain the same
result as if B was defined as a factor (I suppose it comes from
the fact "non-factor arguments a are coerced via factor" according
to the details of the table help). I find it completely normal if
I remember that aggregate first splits the data into subsets and
then compute the table. But then I don't understand why it works
differently with character vectors. Indeed if I use character
vectors, I get the same result as with factors:
------------------------
> df <-
data.frame(A=factor(c("1","1","1","1","0","0","0","0")),B=factor(c("1","0","1","0","0","0","1","0")),C=factor(c("1","0","1","0","0","1","1","1")))
> aggregate(df[1:2],list(df$C),table,simplify = TRUE,drop=TRUE)
Group.1 A.0 A.1 B.0 B.1
1 0 1 2 3 0
2 1 3 2 2 3
> df <-
data.frame(A=factor(c(1,1,1,1,0,0,0,0)),B=factor(c(1,0,1,0,0,0,1,0)),C=factor(c(1,0,1,0,0,1,1,1)))
> aggregate(df[1:2],list(df$C),table,simplify = TRUE,drop=TRUE)
Group.1 A.0 A.1 B.0 B.1
1 0 1 2 3 0
2 1 3 2 2 3
---------------------
Is it possible to precise anything about this behaviour in the
aggregate help since the result is not completely compatible with
the expectation of result we can have according to the table help?
Or would it be possible to have the same results independently of
the vector type? This post was rejected on the R-devel mailing
list so I ask my question here as suggested.
Best regards,
Alain Guillet
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